0.500 atm of NH_3(g) was placed in a reaction vessel maintained at 298 K. When t

Question

0.500 atm of NH_3(g) was placed in a reaction vessel maintained at 298 K. When the system reached equilibrium, the partial pressure of N_2 was found to be 0.011 atm. Calculate the value of K_P at 298K the decomposition reaction 2 NH_3(g) N_2(g) + 3 H_2(g) A. 1.7 times 10^-6 B. 2.3 times 10^-5 C. 3.8 times 10^-6 D. 1.9 times 10^-4. In which reaction would an increase in pressure at constant temperature have no effect on the relative amounts of the substances present in the equilibrium mixture? All substances are gases. (A) 2NO + O_2 2NO_2 + heat (B) heat + Na_2 + O_2 2NO (C) N_2 + 3H_2 2NH_3 + heat (D) 2CO + O_2 2CO_2 + heat The solubility of solid silver chromate, Ag_2CrO_4, which has K_sp equal to 9.0 times 10^-12 at 25 degree C, is determined in water and in two different aqueous solutions. Predict the relative solubility of Ag_2CrO_4 in the three solutions. (A) I = II = III (B) I

Explanation / Answer

Q28

The reaction:

2NH3 = N2 + 3H2

Initially:

P-NH3 = 0.5 atm

P-N2 = 0

P-H2 = 0

In equilibrium

P-NH3 = 0.5 - 2x

P-N2 = 0 +x

P-H2 = 0 +3x

and we know

P-N2 = 0.011 atm

so

P-N2 = 0 +x = 0.011 atm

x = 0.011

substitute

P-NH3 = 0.5 - 2*0.011 = 0.478 atm

P-N2 = 0 +0.011 = 0.011 atm

P-H2 = 0 +3*0.011 = 0.033 atm

so...

Kp = P-N2 * (P-H2)^3 /(NH3)^2

Kp = 0.011 * (0.033 )^3 /(0.478 )^2

Kp = 0.00000173012

Kp = 1.73*10^-6

nearest answer is A

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